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A355878
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Smallest p == 1 (mod 4) such that Q(sqrt(p)) has class number 2n+1.
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2
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5, 229, 401, 577, 1129, 1297, 8101, 11321, 11257, 18229, 7057, 23593, 27689, 8761, 56857, 146077, 63361, 25601, 24337, 55441, 439573, 14401, 32401, 78401, 70969, 69697, 376897, 106537, 41617, 160001, 193601, 57601, 197137, 367721, 414433, 1432813, 444089, 331777
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OFFSET
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0,1
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COMMENTS
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Also smallest odd prime p such that Q(sqrt(p)) has narrow class number (also called form class number) 2n+1.
Conjecture: a(n) > A002148(n) for all n.
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LINKS
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FORMULA
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EXAMPLE
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p = 229 is the smallest odd prime such that Q(sqrt(p)) has class number 3, so a(1) = 229.
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PROG
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(PARI) a(n) = forprime(p=2, oo, if(p%4==1 && qfbclassno(p)==2*n+1, return(p)))
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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